Problem: Find $ \int\dfrac{1}{(x+1)(3x+1)} \,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12\text{ln}|3x+1|-\dfrac12\text{ln}|x+1|+C$ (Choice B) B $\dfrac12\text{ln}|x+1|-\dfrac12\text{ln}|3x+1|+C$ (Choice C) C $\dfrac12\text{ln}|x+1|-\dfrac32\text{ln}|3x+1|+C$ (Choice D) D $\dfrac32\text{ln}|3x+1|-\dfrac12\text{ln}|x+1|+C$
The integrand is a rational function where the degree of the numerator is smaller than the degree of the denominator. Also, the denominator is factorable. This suggests that we should rewrite the integrand using partial fractions: $\int\dfrac{1}{(x+1)(3x+1)}\, dx=\int\left(\dfrac{A}{x+1}+\dfrac{B}{3x+1}\right)\,dx$ Then, we will be able to integrate using the following formula, which is based on the derivative of $\text{ln}(x)$ : $\int\dfrac{1}{mx+n}\,dx=\dfrac1m\cdot\text{ln}|mx+n|+C$ To find $A$ and $B$, we take the following equation and multiply both sides by the common denominator: $\begin{aligned} \dfrac{1}{(x+1)(3x+1)}&=\dfrac{A}{x+1}+\dfrac{B}{3x+1} \\\\ 1&=A(3x+1)+B(x+1) \end{aligned}$ Substituting $x=-1$, we get ${A=-\dfrac12}$. Substituting $x=-\dfrac13$, we get ${B=\dfrac32}$. We can now rewrite the integrand and integrate the expression: $\begin{aligned} &\phantom{=}\int\dfrac{1}{(x+1)(3x+1)} \,dx \\\\ &=\int\left(\dfrac{ A}{x+1}+\dfrac{ B}{3x+1}\right)\,dx \\\\ &=\int\dfrac{{-\dfrac12}}{x+1}+\dfrac{{\dfrac32}}{3x+1}\, dx \\\\ &=-\dfrac12\int\dfrac{1}{x+1}\, dx+\dfrac32\cdot{\dfrac13}\int\dfrac{{3}}{3x+1}\, dx \\\\ &=-\dfrac12\text{ln}|x+1|+\dfrac12\text{ln}|3x+1|+C \\\\ &=\dfrac12\text{ln}|3x+1|-\dfrac12\text{ln}|x+1|+C \end{aligned}$ In conclusion, $ \int\dfrac{1}{(x+1)(3x+1)} \,dx =\dfrac12\text{ln}|3x+1|-\dfrac12\text{ln}|x+1|+C$